如何使用金字塔.response.FileI

Question

我有以下视图代码，试图将zipfile“流式”传输到客户端以供下载：

import os
import zipfile
import tempfile
from pyramid.response import FileIter

def zipper(request):
    _temp_path = request.registry.settings['_temp']
    tmpfile = tempfile.NamedTemporaryFile('w', dir=_temp_path, delete=True)

    tmpfile_path = tmpfile.name

    ## creating zipfile and adding files
    z = zipfile.ZipFile(tmpfile_path, "w")
    z.write('somefile1.txt')
    z.write('somefile2.txt')
    z.close()

    ## renaming the zipfile
    new_zip_path = _temp_path + '/somefilegroup.zip'
    os.rename(tmpfile_path, new_zip_path)

    ## re-opening the zipfile with new name
    z = zipfile.ZipFile(new_zip_path, 'r')
    response = FileIter(z.fp)

    return response

但是，我在浏览器中得到的回答是：

Could not convert return value of the view callable function newsite.static.zipper into a response object. The value returned was .

我想我没有正确使用FileIter。在

---

更新：

由于更新了Michael Merickel的建议，FileIter函数工作正常。但是，仍然存在一个MIME类型错误，该错误出现在客户端（浏览器）上： Resource interpreted as Document but transferred with MIME type application/zip: "http://newsite.local:6543/zipper?data=%7B%22ids%22%3A%5B6%2C7%5D%7D"

为了更好地说明这个问题，我在Github上包含了一个小的.py和{}文件：https://github.com/thapar/zipper-fix

Answer 1

FileIter不是响应对象，就像您的错误消息所说的那样。它是一个可用于响应体的iterable，仅此而已。另外，ZipFile可以接受file对象，这在这里比文件路径更有用。让我们尝试写入tmpfile，然后将该文件指针倒回起始位置，并使用它在不进行任何花哨重命名的情况下写出。在

import os
import zipfile
import tempfile
from pyramid.response import FileIter

def zipper(request):
    _temp_path = request.registry.settings['_temp']
    fp = tempfile.NamedTemporaryFile('w+b', dir=_temp_path, delete=True)

    ## creating zipfile and adding files
    z = zipfile.ZipFile(fp, "w")
    z.write('somefile1.txt')
    z.write('somefile2.txt')
    z.close()

    # rewind fp back to start of the file
    fp.seek(0)

    response = request.response
    response.content_type = 'application/zip'
    response.app_iter = FileIter(fp)
    return response

根据文档，我将NamedTemporaryFile上的模式改为'w+b'，以允许文件写入和读取。在